# general commutativity

If the binary operation  $\cdot$” on the set $S$ is commutative   , then for each  $a_{1},a_{2},\ldots,a_{n}$ in $S$ and for each permutation  $\pi$ on  $\{1,\,2,\,\ldots,\,n\}$,  one has

 $\displaystyle\prod_{i=1}^{n}a_{\pi(i)}\;=\;\prod_{i=1}^{n}a_{i}.$ (1)

Proof.  If  $n=1$,  we have nothing to prove.  Make the induction hypothesis, that (1) is true for  $n=m\!-\!1$.  Denote

 $\pi^{-1}(m)\;=\;k,\quad\mbox{i.e.}\quad\pi(k)=m.$

Then

 $\prod_{i=1}^{m}a_{\pi(i)}\;=\;\prod_{i=1}^{k-1}a_{\pi(i)}\cdot a_{\pi(k)}\cdot% \prod_{i=1}^{m-k}a_{\pi(k+i)}\;=\;\left(\prod_{i=1}^{k-1}a_{\pi(i)}\cdot\prod_% {i=1}^{m-k}a_{\pi(k+i)}\right)\cdot a_{m},$

where $a_{m}$ has been moved to the end by the induction hypothesis.  But the product   in the parenthesis, which exactly the factors $a_{1},a_{2},\ldots,a_{m-1}$ in a certain , is also by the induction hypothesis equal to $\prod_{i=1}^{m-1}a_{i}$.  Thus we obtain

 $\prod_{i=1}^{m}a_{\pi(i)}\;=\;\prod_{i=1}^{m-1}a_{i}\cdot a_{m}\;=\;\prod_{i=1% }^{m}a_{i},$

whence (1) is true for  $n=m$.

Note.  There is mentionned in the Remark of the entry “http://planetmath.org/node/2148commutativity” a more general notion of commutativity.

Title general commutativity GeneralCommutativity 2014-05-10 21:59:41 2014-05-10 21:59:41 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 20-00 CommutativeLanguage GeneralAssociativity AbelianGroup2